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I am reading Optics by Eugene Hecht, and I am confused about the author's derivation of the differential wave equation:

To relate the space and time dependencies of $\psi(x, t)$, take the partial derivative of $\psi(x, t) = f(x')$ with respect to $x$, holding $t$ constant. Using $x' = x \mp vt$, and inasmuch as

$$\dfrac{\partial{\psi}}{\partial{x}} = \dfrac{\partial{f}}{\partial{x}}$$

$$\dfrac{\partial{\psi}}{\partial{x}} = \dfrac{\partial{f}}{\partial{x'}} \dfrac{\partial{x'}}{\partial{x}} = \dfrac{\partial{f}}{\partial{x'}} \tag{2.8}$$

because $$\dfrac{\partial{x'}}{\partial{x}} = \dfrac{\partial{x \mp vt}}{\partial{x}} = 1$$

Holding $x$ constant, the partial derivative with respect to time is

$\dfrac{\partial{\psi}}{\partial{t}} = \dfrac{\partial{f}}{\partial{x'}} \dfrac{\partial{x'}}{\partial{t}} = \dfrac{\partial{f}}{\partial{x'}} (\mp v) = \mp v \dfrac{\partial{f}}{\partial{x'}} \tag{2.9}$

Combining Eqs. (2.8) and (2.9) yields

$$\dfrac{\partial{\psi}}{\partial{t}} = \mp v \dfrac{\partial{\psi}}{\partial{x}}$$

This says that the rate of change of $\psi$ with $t$ and with $x$ are equal, to within a multiplicative constant, as shown in Fig. 2.5. The second partial derivatives of Eqs. (2.8) and (2.9) are

$$\dfrac{\partial^2{\psi}}{\partial{x}^2} = \dfrac{\partial^2{f}}{\partial{x'}^2} \tag{2.10}$$

and $$\dfrac{\partial^2{\psi}}{\partial{t}^2} = \dfrac{\partial}{\partial{t}} \left( \mp v \dfrac{\partial{f}}{\partial{x'}} \right) = \mp v \dfrac{\partial}{\partial{x'}} \left( \dfrac{\partial{f}}{\partial{t}} \right)$$

Since $$\dfrac{\partial{\psi}}{\partial{t}} = \dfrac{\partial{f}}{\partial{t}}$$

$$\dfrac{\partial^2{\psi}}{\partial{t}^2} = \mp \dfrac{\partial}{\partial{x'}} \left( \dfrac{\partial{\psi}}{\partial{t}} \right)$$

It follows, using Eqn. (2.9), that $$\dfrac{\partial^2{\psi}}{\partial{t}^2} = v^2 \dfrac{\partial^2{f}}{\partial{x'}^2}$$

I don't see how

$$\dfrac{\partial{\psi}}{\partial{t}} = \dfrac{\partial{f}}{\partial{t}} \tag{1}$$

We actually found that $\dfrac{\partial{\psi}}{\partial{t}} = \mp v \dfrac{\partial{\psi}}{\partial{x}}$, so I have no idea where the author got $\dfrac{\partial{\psi}}{\partial{t}} = \dfrac{\partial{f}}{\partial{t}}$ from?

I would greatly appreciate it if people could please take the time to clarify this.

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  • $\begingroup$ $$\dfrac{\partial{\psi}}{\partial{t}} = \dfrac{\partial{\psi}}{\partial{x}}\dfrac{\partial{x}}{\partial{t}}+\dfrac{\partial{\psi}}{\partial{t}}\dfrac{\partial{t}}{\partial{t}} = \dfrac{\partial{f}}{\partial{t}}$$ but $$\dfrac{\partial{x}}{\partial{t}}=0$$ $\endgroup$ – Nosrati Sep 20 '18 at 14:52
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$$ \psi(x, t) = f(\bar{x}) \to \frac{\partial}{\partial t}\psi(x, t) = \frac{\partial}{\partial t}f(\bar{x}) $$ this is true. The fact that $\frac{\partial}{\partial t}f(\bar{x})$ also equals something else is secondary.

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